Final Assignment

**by**

__PART
A__

Consider any triangle ABC. Select point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points, D, E, and F respectively.

Explore
(AF)(BD)(CE) and (BF)(CD)(AE) for various triangles and various locations of P.

To
begin, we will pick various locations of point P to see what affect it has on
the relationship of the segments of the sides of the triangle. If we multiply the length of the
following sides together, what occurs?

To
move A, B, C, and P around, CLICK HERE!

When
point P is the centroid, triangle ABC is divided into six triangles with equal
area. Points D, E, and F are
midpoints of ÆABC and point P is the point of concurrency. Therefore, if we multiply the length of
the following sides together, does the same thing happen?

CLICK HERE to move points A, B, and C
around the centroid, P.

__PART
B__

A conjecture
drawn by this manipulation is that (AF)(BD)(CE) = (BF)(CD)(AE), or .

To
prove this, letÕs start by extending the lines BE and CF until they meet at GH,
which is through A and parallel to BC.

There
are several pairs of similar triangles: ÆAFH, & ÆBCF, and ÆAEG, & ÆBCE,
also ÆAGP, & ÆBDP, as well as ÆCDP, & ÆAHP.

From
these, we can derive the following proportions:

, , , and .

Since
and , then .

Therefore,
.

By
multiplying three of these identities together, we get:

.

Therefore,
we find that:

.

Therefore, if the lines AD, BE and CF intersect at a single point P,
the identity does hold.

Does
this identity hold regardless of where P is located?

To
use GSP, CLICK HERE!

__PART
C__

Looking
back at triangle ABC and arbitrary point P inside the triangle, we can see
through the GSP construction that the ratio of the areas of triangle ABC and
triangle DEF is always greater than or equal to 4.

To
move point P to different locations inside the triangle and watch the value of
the ratio, CLICK HERE!

The
ratio is equal to four only when D, E, and F are the midpoints of triangle ABC
and point P is concurrent with the centroid of the triangle. We then have a medial triangle formed by
the points. To move A, B, and C around the centroid, CLICK
HERE!